( $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}, \log 4=0.6021$ આપેલ છે.)
\( \log \left(\frac{4}{1}\right)=\frac{E_a}{2.303 R}\left(\frac{1}{300}-\frac{1}{330}\right) \)
\( E_a=\frac{(\log (4)) \times 2.303 \times 8.314 \times 300 \times 330}{30} \)
\( =3.804 \times 10^4 \mathrm{~J} / \mathrm{mol} \)
\( =38.04 \mathrm{~kJ} / \mathrm{mol}\)
(નજીકના પૂર્ણાંકમાં રાઉન્ડ ઑફ) (ધારી લો : $\ln 10=2.303, \ln 2=0.693$)